Does this inequality hold for the cumulant generating function?

Question

Suppose a random variable $X$ is zero-mean and the cumulant generating function is $$ K\left( t \right) =\log \mathbb{E}[e^{tX}]. $$ Given any positive constant $\tau > 0$, does this inequality $$ \left( \tau -t \right) K^\prime\left( t \right) +K\left( t \right) \ge \frac{t}{\tau}K\left( \tau \right) $$ hold for all $ 0 \le t \le \tau$? Here $K^\prime$ is the derivative function of $K$.

Answer 1

This is not true in general. Indeed, let $X$ be a zero-mean random variable (r.v.) such that $Ee^{tX}<\infty$ for $t\in[0,\tau)$ but $Ee^{\tau X}=\infty$. Then for all $t\in(0,\tau)$ the left-hand side of the conjectured inequality will be finite but the right-hand side of the inequality will be $\infty$.

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It is quite easy to modify the above example by approximation so as to make the r.v. $X$ bounded. E.g., let $X:=X_n:=\min(Y,n)-\mu_n$, where $Y$ is an exponential r.v. with mean $1$ and $\mu_n:=E\min(Y,n)=1-2 e^{-n} n-e^{-2 n}\to1$ (as $n\to\infty$). Then the r.v. $X$ is bounded and $$K(t)=K_n(t)=\ln\frac{t e^{(t-1)n}-1}{t-1}-\mu_n t$$ for real $t\ne1$, with $K(1)=K_n(1)=\ln(n+1)-\mu_n$. So, letting $n\to\infty$, for real $t\in(0,1)$ we get $K(t)=K_n(t)\to\ln\frac1{1-t}-t$ and $K'(t)=K'_n(t)\to\frac t{1-t}$, so that the left-hand side of the conjectured inequality goes to a finite limit, whereas the limit of the right-hand side of the conjectured inequality for $\tau=1$ is $\lim_{n\to\infty}tK_n(1)=\lim_{n\to\infty}t(\ln(n+1)-\mu_n)=\infty$. So, for each $t\in(0,1)$ and all large enough $n$ the conjectured inequality will fail to hold.